# How do I find the antiderivative of sec^3 x tan^4 x dx?

##### 1 Answer
Mar 23, 2015

I'm going to guess that you'll need trigonometric power reduction formulas and integration by parts and perhaps integration by substitution.
I'm guessing this based on an answer by of:

$\frac{1}{96} \left[\frac{1}{8} \left(78 \sin x - 47 \sin 3 x + 3 \sin 5 x\right) {\sec}^{6} x - 6 \ln \left(\cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right)\right) + 6 \ln \left(\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)\right)\right]$.

from WolframAlpha.

WolframAlpha is not particularly efficient though.

For example: $- 6 \ln \left(\cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right)\right) + 6 \ln \left(\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)\right) = - 6 \ln \left(\frac{1 - \sin x}{\cos} x\right)$

So the answer could be written:

$\frac{1}{96} \left[\frac{1}{8} \left(78 \sin x - 47 \sin 3 x + 3 \sin 5 x\right) {\sec}^{6} x - 6 \ln \left(\frac{1 - \sin x}{\cos} x\right)\right]$.

Good luck.